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1 fL fL
~ I f (x ) dx = I dx = L ; L J-L J0
1 fL mnx
f (x ) cos- dx =
= 0, m = 0
mn x cos dx
Similarly, from Eq. (3),
1 fL m nx
lJ f (x ) sm dx =
m n x sin dx
(1 cos mn)
0, m even;
2L /m n, m odd.
10.3 The Fourier Convergence Theorem
L 2L ( nx 1 3nx 1 5nx
f (x) = - + sin + - sin - + - sin - + ?
2 n \ L 3 L 5 L
L 2L sin(m nx / L)
2 n m
m = 1,3,5,...
L 2L sin(2n 1)nx/L _
= T + (6)
2 n f 2« 1
n = 1
At the points x = 0, ±nL, where the function f in the example is not continuous, all terms in the series after the first vanish and the sum is L /2. This is the mean value of the limits from the right and left, as it should be. Thus we might as well define f at these points to have the value L/2. If we choose to define it otherwise, the series still gives the value L /2 at these points, since none of the preceding calculations is altered in any detail; it simply does not converge to the function at those points unless f is defined to have this value. This illustrates the possibility that the Fourier series corresponding to a function may not converge to it at points of discontinuity unless the function is suitably defined at such points.
The manner in which the partial sums
L 2L / nx 1 . (2n 1)nx \
s«(x) = 2 + it (sin t + ??? + sms,nLj n = 12
of the Fourier series (6) converge to f (x) is indicated in Figure 10.3.3, where L has been chosen to be one and the graph of s8 (x) is plotted. The figure suggests that at points where f is continuous the partial sums do approach f (x) as n increases. However, in the neighborhood of points of discontinuity, such as x = 0 and x = L, the partial sums do not converge smoothly to the mean value. Instead they tend to overshoot the mark at each end of the jump, as though they cannot quite accommodate themselves to the sharp turn required at this point. This behavior is typical of Fourier series at points of discontinuity, and is known as the Gibbs4 phenomenon.
\A Amaaaa/1 - n = 8
\/VVVVV\/ i (/vvvvvv
-2 -1 1 2 x
FIGURE 10.3.3 The partial sum s8 (x) in the Fourier series, Eq. (6), for the square wave.
Additional insight is attained by considering the error en (x) = f (x) sn (x). Figure
10.3.4 shows a plot of |en (x) | versus x for n = 8 and for L = 1. The least upper bound of |e8(x)| is 0.5 and is approached as x ^ 0 and as x ^ 1. As n increases, the error
4The Gibbs phenomenon is named after Josiah Willard Gibbs (1839-1903), who is better known for his work on vector analysis and statistical mechanics. Gibbs was professor of mathematical physics at Yale, and one of the first American scientists to achieve an international reputation. Gibbs phenomenon is discussed in more detail by Carslaw (Chapter 9).
Chapter 10. Partial Differential Equations and Fourier Series
decreases in the interior of the interval [where f (x) is continuous] but the least upper bound does not diminish with increasing n. Thus one cannot uniformly reduce the error throughout the interval by increasing the number of terms.
Figures 10.3.3 and 10.3.4 also show that the series in this example converges more slowly than the one in Example 1 in Section 10.2. This is due to the fact that the coefficients in the series (6) are proportional only to 1/(2n 1).
PROBLEMS In each of Problems 1 through 6 assume that the given function is periodically extended outside the original interval.
(a) Find the Fourier series for the extended function.
(b) Sketch the graph of the function to which the series converges for three periods.
1. f (x ) =
3. f (x) =
5. f(x) =
1, 1 < x < 0,
0 < x < 1
2. f (x) =
0, n < x < 0,
x, 0 < x < n
L + x, L < x < 0,
0 < x < L
0, n < x < n/2,
1, n/2 < x < n/2,
0, n/2 < x < n
4. f (x) = 1 x2, 1 < x < 1
0, 1 < x < 0,
6. f(x) = x2, 0 < x < 1
In each of Problems 7 through 12 assume that the given function is periodically extended outside the original interval.
(a) Find the Fourier series for the given function.
(b) Let en (x) = f (x) sn (x). Find the least upper bound or the maximum value (if it exists) of len(x)| for n = 10, 20, and 40.
(c) If possible, find the smallest n for which |en (x)| < 0.01 for all x.
? 7. f (x) = P
n < x < 0, 0 < x < n;
f (x + 2n) = f (x) (see Section 10.2, Problem 15)
10.3 The Fourier Convergence Theorem
? 8. f (x) = j! + 1 ^ < x < 1; f (x + 2) = f (x) (see Section 10.2, Problem 16)
? 9. f (x) = x, 1 < x < 1; f (x + 2) = f (x) (see Section 10.2, Problem 20)
,,, , |x + 2, 2< x < 0, r, , ,,, , (see Section 10.2,
? I0. f (x) = 12 2x, 0 < x < 2; f (x + 4) = f (x) Problem 22)
? 11. f (x) = |02 0 < x < 0; f (x + 2) = f (x) (see Problem 6)
? 12. f (x) = x x3, 1 < x < 1; f (x + 2) = f (x)
Periodic Forcing Terms. In this chapter we are concerned mainly with the use of Fourier series
to solve boundary value problems for certain partial differential equations. However, Fourier series are also useful in many other situations where periodic phenomena occur. Problems 13 through 16 indicate how they can be employed to solve initial value problems with periodic forcing terms.